3

我很难找到一种有效的方法来根据其标签在数据库中选择一行，并返回与该行关联的所有其他标签。当我使用不返回行的所有标签的查询时，大约需要 0.001 秒。我的初始方案更加规范化，并且有另一个用于标签标签的表，但它最终需要几秒钟才能完成一个查询，所以我结束了删除该表并使其不那么规范化，但即使这个解决方案似乎也很慢。

SELECT c.*
FROM collections c,
     tags t
WHERE t.collection_id=c.id
  AND (t.name IN ("foo",
                  "bar"))
GROUP BY c.id HAVING COUNT(t.id)=2 LIMIT 10

现在我无法想出一种有效的方法来获取该元素的所有其他标签而不会变慢。我目前的解决方案慢了大约 10 倍，需要 0.01 秒才能完成，而且我觉得它的扩展性不好（而且我觉得它很丑）。

SELECT c.*,
       GROUP_CONCAT(t1.name) AS tags
FROM collections c,
     tags t,
     tags t1
WHERE t1.collection_id = c.id
  AND t.collection_id=c.id
  AND (t.name IN ("foo",
                  "bar"))
GROUP BY c.id HAVING COUNT(t.id)=2 LIMIT 10

实际上是否有一种有效或至少更有效的方法来实现这一目标？非常感谢您对此提供任何建议或提示！

4

2 回答 2

0

好的。考虑以下...

DROP TABLE IF EXISTS ingredients;

CREATE TABLE ingredients 
(ingredient_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,ingredient VARCHAR(30) NOT NULL UNIQUE
);

INSERT INTO ingredients (ingredient_id, ingredient) VALUES
(1, 'Macaroni'),
(2, 'Cheese'),
(3, 'Beans'),
(4, 'Toast'),
(5, 'Jam'),
(6, 'Jacket Potato'),
(7, 'Peanut Butter');


DROP TABLE IF EXISTS recipes;

CREATE TABLE recipes 
(recipe_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,recipe VARCHAR(50) NOT NULL UNIQUE
);

INSERT INTO recipes (recipe_id, recipe) VALUES
(1, 'Macaroni & Cheese'),
(2, 'Cheese on Toast'),
(3, 'Beans on Toast'),
(4, 'Cheese & Beans on Toast'),
(5, 'Toast & Jam'),
(6, 'Beans & Macaroni'),
(9, 'Beans on Jacket Potato'),
(10, 'Cheese & Beans on Jacket Potato'),
(12, 'Peanut Butter on Toast');

DROP TABLE IF EXISTS recipe_ingredient;

CREATE TABLE recipe_ingredient 
(recipe_id INT NOT NULL
,ingredient_id INT NOT NULL
,PRIMARY KEY (recipe_id,ingredient_id)
);

INSERT INTO recipe_ingredient (recipe_id, ingredient_id) VALUES
(1, 1),
(1, 2),
(2, 2),
(2, 4),
(3, 3),
(3, 4),
(4, 2),
(4, 3),
(4, 4),
(5, 4),
(5, 5),
(6, 1),
(6, 3),
(9, 3),
(9, 6),
(10, 2),
(10, 3),
(10, 6),
(12, 4),
(12, 7);

SELECT r.*
      , GROUP_CONCAT(CASE WHEN i.ingredient IN ('Cheese','Beans') THEN i.ingredient END) i
      , GROUP_CONCAT(CASE WHEN i.ingredient NOT IN('Cheese','Beans') THEN i.ingredient END) o 
   FROM recipes r 
   LEFT 
   JOIN recipe_ingredient ri 
     ON ri.recipe_id = r.recipe_id 
   LEFT 
   JOIN ingredients i 
     ON i.ingredient_id = ri.ingredient_id 
  GROUP 
     BY recipe_id;

+-----------+---------------------------------+--------------+---------------------+
| recipe_id | recipe                          | i            | o                   |
+-----------+---------------------------------+--------------+---------------------+
|         1 | Macaroni & Cheese               | Cheese       | Macaroni            |
|         2 | Cheese on Toast                 | Cheese       | Toast               |
|         3 | Beans on Toast                  | Beans        | Toast               |
|         4 | Cheese & Beans on Toast         | Cheese,Beans | Toast               |
|         5 | Toast & Jam                     | NULL         | Toast,Jam           |
|         6 | Beans & Macaroni                | Beans        | Macaroni            |
|         9 | Beans on Jacket Potato          | Beans        | Jacket Potato       |
|        10 | Cheese & Beans on Jacket Potato | Cheese,Beans | Jacket Potato       |
|        12 | Peanut Butter on Toast          | NULL         | Toast,Peanut Butter |
+-----------+---------------------------------+--------------+---------------------+

相同的小提琴：http ://www.sqlfiddle.com/#!2/45aa0/1

于 2013-10-18T10:31:03.967 回答
0

让它使用显式连接语法（这不应该对性能产生影响，因为 MySQL 应该设法优化它）

SELECT c.*,
       GROUP_CONCAT(t1.name) AS tags
FROM collections c
INNER JOIN tags t ON t.collection_id = c.id
INNER JOIN tags t1 ON t1.collection_id = c.id
WHERE t.name IN ("foo", "bar")
GROUP BY c.id 
HAVING COUNT(t.id) = 2 
LIMIT 10

可能值得为您检查的每个标签做一个单独的 INNER JOIN，这样就不需要 HAVING：-

SELECT c.*,
       GROUP_CONCAT(t1.name) AS tags
FROM collections c
INNER JOIN tags t ON t.collection_id = c.id AND t.name = "foo"
INNER JOIN tags t0 ON t.collection_id = c.id AND t0.name = "bar"
INNER JOIN tags t1 ON t1.collection_id = c.id
GROUP BY c.id 
LIMIT 10

但是，您的原始查询看起来不错，因此可能是索引问题。

于 2013-10-18T10:55:21.250 回答